The matrix for rotating an ordinary vector by $\phi$ around the $z$-axis is
$$
\mathbf{R}(\phi) \equiv\left(\begin{array}{ccc}
\cos \phi & -\sin \phi & 0 \\
\sin \phi & \cos \phi & 0 \\
0 & 0 & 1
\end{array}\right)
$$
By considering the form taken by $\mathbf{R}$ for infinitesimal $\phi$ calculate from $\mathbf{R}$ the matrix $\mathcal{J}_{z}$ that appears in $\mathbf{R}(\phi)=\exp \left(-\mathrm{i} \mathcal{J}_{z} \phi\right)$. Introduce new coordinates $u_{1} \equiv(-x+\mathrm{i} y) / \sqrt{2}, u_{2}=z$ and $u_{3} \equiv(x+\mathrm{i} y) / \sqrt{2}$. Write down the matrix $\mathbf{M}$ that appears in $\mathbf{u}=\mathbf{M} \cdot \mathbf{x}$ [where $\mathbf{x} \equiv(x, y, z)]$ and show that it is unitary. Then show that
$$
\mathcal{J}_{z}^{\prime} \equiv \mathbf{M} \cdot \mathcal{J}_{z} \cdot \mathbf{M}^{\dagger}
$$
is identical with $S_{z}$ in the set of spin-one Pauli analogues
$$
S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right), \quad S_{y}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & -\mathrm{i} & 0 \\
\mathrm{i} & 0 & -\mathrm{i} \\
0 & \mathrm{i} & 0
\end{array}\right), \quad S_{z}=\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array}\right)
$$